ChemistryChemistry

ThermodynamicsThermodynamics

I. Enthalpy: Endothermic & I. Enthalpy: Endothermic & Exothermic ReactionsExothermic Reactions

A. NRG A. NRG can be thought ofcan be thought of as a reactant or as a reactant or product: product:

NRG + Reactant NRG + Reactant →→ Product Product

(Endothermic)(Endothermic)

Reactant Reactant → → Product + NRGProduct + NRG

(Exothermic)(Exothermic)

B. TerminologyB. Terminology

1. Enthalpy (1. Enthalpy (HH): Potential Energy associated ): Potential Energy associated with a substance (Wile)with a substance (Wile)

- Kind of right – we’ll go with it because - Kind of right – we’ll go with it because it’s it’s simplesimple

2. Enthalpy of reaction (∆2. Enthalpy of reaction (∆HH): The difference ): The difference between the enthalpies of the products minus between the enthalpies of the products minus the enthalpies of the reactantsthe enthalpies of the reactants

∆∆H = H H = H (products) – (products) – HH (reactants) (reactants)

3. Units of ∆3. Units of ∆HH: :

NRG units (calories, Joules)NRG units (calories, Joules)

1.000 cal = 4.184 Joule1.000 cal = 4.184 Joule

4. Signs:4. Signs:

∆∆HH is + for endothermic is + for endothermic

∆∆HH is - for exothermic is - for exothermic

5. Example:5. Example:

When CHWhen CH4 4 is burned, the enthalpy of the is burned, the enthalpy of the

reaction is -802.2 kJ/mol. Write a balanced reaction is -802.2 kJ/mol. Write a balanced chemical equation representing this chemical equation representing this process. Include NRG in your equation.process. Include NRG in your equation.

OYOOYO

13.1 Write balanced chemical equations for 13.1 Write balanced chemical equations for each of the following processes. Include each of the following processes. Include NRG as either a reactant or product.NRG as either a reactant or product.

A) Formation of HA) Formation of H22SOSO44; ∆; ∆HH = -814 kJ = -814 kJ

B) Decomposition of KB) Decomposition of K33POPO44; ∆; ∆HH = 2,030 kJ = 2,030 kJ

C) Reaction of HC) Reaction of H22COCO33 & KOH; releases 21 & KOH; releases 21

kcalskcals

II. Determining ∆II. Determining ∆HH for a Chemical for a Chemical Reaction by CalorimetryReaction by Calorimetry

Remember q = mc∆T ?Remember q = mc∆T ?

q is negative when NRG is releasedq is negative when NRG is released

q is positive when NRG is absorbedq is positive when NRG is absorbed

∆∆HH is equal (or very close) to q is equal (or very close) to qrxnrxn

III. Determining ∆III. Determining ∆HH using Standard using Standard Enthalpy of FormationsEnthalpy of Formations

A. The definitionA. The definition

Standard Enthalpy of Formation (∆Standard Enthalpy of Formation (∆HH ffoo): ):

the ∆the ∆HH of a formation reaction at standard of a formation reaction at standard conditionsconditions

““o”o” means means Standard ConditionsStandard Conditions (25 (25ooC, 1.00 atm)C, 1.00 atm)

-- different from STP (0.0 -- different from STP (0.0 ooC, 1.00 atm)C, 1.00 atm)

““ff” means formation” means formation

We can use ∆We can use ∆HH ffo o for reactants and products for reactants and products

instead of instead of HH (which we can’t measure) (which we can’t measure)

So, we said So, we said ∆∆HH = H= H (products) – (products) – HH (reactants) (reactants)

can’t measurecan’t measure can’t measure can’t measure

We’ll use:We’ll use:B. The equationB. The equation

∆∆HHrxnrxn oo = ∑ = ∑ ∆∆HH ffo o (prod) - (prod) - ∑ ∑ ∆∆HH ff

o o (reac.)(reac.) ((∑ ∑ means “sum of”)means “sum of”)

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C. Some Standard Enthalpies of FormationC. Some Standard Enthalpies of Formation

-By convention, the standard enthalpy of -By convention, the standard enthalpy of formation for any element = 0formation for any element = 0

- Note that the phase is important!- Note that the phase is important!

- Note that units are kJ/mole- Note that units are kJ/mole

D. Exothermic or Endothermic?D. Exothermic or Endothermic?Using ∆Using ∆HHrxnrxn oo = ∑ = ∑ ∆∆HH ff

o o (prod) - (prod) - ∑ ∑ ∆∆HH ffo o (reac.)(reac.)

1. What is1. What is ∆∆HHrxnrxn oo of the following reaction?of the following reaction?

2NO(g) + O2NO(g) + O22(g) (g) →→ 2NO 2NO22(g)(g)

2. Some gasoline companies put a small 2. Some gasoline companies put a small amount of liquid ethyl alcohol (Camount of liquid ethyl alcohol (C22HH66O) into O) into

their gasoline. What is the ∆their gasoline. What is the ∆HHrxnrxn oo ??

OYO’sOYO’s

13.2 When you use a pocket cigarette lighter, you 13.2 When you use a pocket cigarette lighter, you are burning gaseous butane (Care burning gaseous butane (C44HH1010). Assuming it ). Assuming it is complete combustion, what is ∆is complete combustion, what is ∆HHoo for this for this reaction?reaction?

13.3 In order to make the shells that house and 13.3 In order to make the shells that house and protect them, many shellfish take solid lime protect them, many shellfish take solid lime (CaO) from the ocean floor and react it with (CaO) from the ocean floor and react it with gaseous carbon dioxide that has been exhaled gaseous carbon dioxide that has been exhaled by marine organisms. The resulting calcium by marine organisms. The resulting calcium carbonate is the major component of their shell. carbonate is the major component of their shell. What is the ∆What is the ∆HHoo for this reaction? for this reaction?

IV. Applying Enthalpy to IV. Applying Enthalpy to StoichiometryStoichiometry

Example 1:Example 1:

The ∆The ∆HHoorxn rxn for the combustion of Cfor the combustion of C22HH66O is O is

-1236 kJ. How much energy can be -1236 kJ. How much energy can be produced by burning 150.0 g of ethyl produced by burning 150.0 g of ethyl alcohol in excess oxygen?alcohol in excess oxygen?

Example 2:Example 2:

One way of producing butane (COne way of producing butane (C44HH1010) for ) for lighters is to force ethane (Clighters is to force ethane (C22HH66) to react ) to react with itself:with itself:

2C2C22HH66(l) (l) →→ C C44HH1010(g) + H(g) + H22(g)(g)

If you have 1.23 kg of ethane and would like If you have 1.23 kg of ethane and would like to turn it all into butane, how many kJ of to turn it all into butane, how many kJ of energy would be necessary?energy would be necessary?

OYO’sOYO’s13.4 Ethane gas (C13.4 Ethane gas (C22HH66) can be used for auto fuel ) can be used for auto fuel

because it releases a lot of NRG when burned. because it releases a lot of NRG when burned. How many kJ of NRG are released when 250.0 How many kJ of NRG are released when 250.0 g of ethane are burned?g of ethane are burned?

13.5 When mined, coal has a small amount of 13.5 When mined, coal has a small amount of sulfur in it. When the coal is burned in a power sulfur in it. When the coal is burned in a power plant, SOplant, SO33(g) is produced. When the SO(g) is produced. When the SO33(g) (g) reacts with water in the air, the following reaction reacts with water in the air, the following reaction occurs:occurs:

HH22O(g) + SOO(g) + SO33(g) (g) →→ H H22SOSO44(l)(l)The sulfuric acid falls to earth as acid rain. How The sulfuric acid falls to earth as acid rain. How

much NRG is produced when 50.0 g of Hmuch NRG is produced when 50.0 g of H22SOSO44 are produced in this way?are produced in this way?

V. NRG DiagramsV. NRG DiagramsA. Exothermic ReactionA. Exothermic Reaction

CC22HH66O(l) + HO(l) + H22(g) (g) →→ C C22HH66(g) + H(g) + H22O(g)O(g)

Potential NRG = ∆Potential NRG = ∆HH

Reaction Coordinate = Rxn progressReaction Coordinate = Rxn progress

Intermediate Stage = “Activated Complex”Intermediate Stage = “Activated Complex”

B. Activation NRG: The NRG necessary to B. Activation NRG: The NRG necessary to start a chemical reactionstart a chemical reaction

AllAll reactions require activation NRG reactions require activation NRG

C. NRG Diagram for an C. NRG Diagram for an Endothermic RxnEndothermic Rxn

C. Using NRG DiagramsC. Using NRG Diagrams

Which is most exothermic? ∆Which is most exothermic? ∆HH??

Which is easiest to get started?Which is easiest to get started?

Which is endothermic? ∆Which is endothermic? ∆HH??

OYOOYO13.6 You perform 3 experiments:13.6 You perform 3 experiments:

Experiment A: hard to start, gets very hotExperiment A: hard to start, gets very hot

Experiment B: easy to start, gets only Experiment B: easy to start, gets only slightly warmslightly warm

Draw NRG diagrams for both. (Label the y Draw NRG diagrams for both. (Label the y and x axis – but you will have no numbers)and x axis – but you will have no numbers)

VI. EntropyVI. Entropy

A. The definition:A. The definition:

Entropy Entropy ((SS)):: a measure of how spread out a measure of how spread out the NRG is in any systemthe NRG is in any system

1. Units: J/mol K 1. Units: J/mol K

2. Can measure absolute entropy2. Can measure absolute entropy

(3(3rdrd Law of Thermodynamics: a Law of Thermodynamics: a perfect crystal has 0 perfect crystal has 0 SS at 0.00 K) at 0.00 K)

3. Factors that affect entropy3. Factors that affect entropy

a. Phases of mattera. Phases of matter

solid – leastsolid – least

liquid – nextliquid – next

gas – mostgas – most

b. Entropy increases with temperatureb. Entropy increases with temperature

c. Entropy increases with increasing c. Entropy increases with increasing mattermatter

B. Determining whether entropy B. Determining whether entropy increases or decreasesincreases or decreases

Is ∆Is ∆S positive or negative?S positive or negative?

Zn(s) + HCl(aq) Zn(s) + HCl(aq) →→ ZnCl ZnCl22(aq) + H(aq) + H22(g)(g)

NN22(g) + 3H(g) + 3H22(g) (g) →→ 2NH 2NH33(g)(g)

2NH2NH44NONO33(s) + Ba(OH)(s) + Ba(OH)22*8H*8H22O(s) O(s) →→

Ba(NOBa(NO33))22(aq) + 2NH(aq) + 2NH44OH(aq) + 8HOH(aq) + 8H22O(l)O(l)

OYO’sOYO’s13.7 How does the entropy of a system 13.7 How does the entropy of a system

change for each of the following change for each of the following processes?processes?

a. A solid meltsa. A solid melts

b. A liquid freezesb. A liquid freezes

c. A liquid boilsc. A liquid boils

d. A vapor is converted to a solidd. A vapor is converted to a solid

e. Urea dissolves in watere. Urea dissolves in water

13.8 Determine the sign of ∆S for the 13.8 Determine the sign of ∆S for the following:following:

a. Ca. C22HH44OO22(aq) + NaHCO(aq) + NaHCO33(s) → (s) →

NaCNaC22HH33OO22(aq) + H(aq) + H22O(l) + COO(l) + CO22(g)(g)

b. 2Hb. 2H22OO22(g) → 2H(g) → 2H22O(g) + OO(g) + O22(g)(g)

c. 2AgNOc. 2AgNO33(aq) + Mg(OH)(aq) + Mg(OH)22(aq) →(aq) →

2AgOH(s) + Mg(NO2AgOH(s) + Mg(NO33))22(aq)(aq)

C. Determining ∆C. Determining ∆SSoo for a Chemical for a Chemical RxnRxn

All substances have a positive entropyAll substances have a positive entropy

(except for a perfect crystalline solid at 0 K)(except for a perfect crystalline solid at 0 K)

SS has the units J/mol K has the units J/mol K

(notice the units are J instead of kJ – (notice the units are J instead of kJ – entropy NRG is entropy NRG is veryvery small comparatively) small comparatively)

∆∆SSrxnrxn oo = ∑ = ∑ ∆∆SS ffo o (prod) - (prod) - ∑ ∑ ∆∆SS ff

o o (reac.)(reac.)

OYOOYO

13.9 Calculate ∆13.9 Calculate ∆SSoo for the following:for the following:

CaO(s) + COCaO(s) + CO22(g) (g) →→ CaCO CaCO33(s)(s)

VII. 2VII. 2ndnd Law of Thermodynamics Law of ThermodynamicsA. Definition: A. Definition:

1. The entropy of the universe increases or remains the same for 1. The entropy of the universe increases or remains the same for spontaneous processesspontaneous processes2. Energy of all types spontaneously flows from being more 2. Energy of all types spontaneously flows from being more localized or concentrated to becoming more dispersed or spread localized or concentrated to becoming more dispersed or spread out, if not hindered.out, if not hindered.

∆∆SSuniverseuniverse≥ 0≥ 0

B. Spontaneous: a reaction that occurs naturally under certain B. Spontaneous: a reaction that occurs naturally under certain circumstancescircumstances

One of the main reasons chemists study thermodynamics is so they One of the main reasons chemists study thermodynamics is so they are able to predict whether a reaction will be spontaneous!are able to predict whether a reaction will be spontaneous!

Spontaneous ProcessesSpontaneous Processes

- Waterfall runs downhill, not up- Waterfall runs downhill, not up- Sugar lump dissolves in water, sugar lump - Sugar lump dissolves in water, sugar lump

doesn’t reappeardoesn’t reappear- Water freezes below 0- Water freezes below 0ooC; ice melts above 0C; ice melts above 0ooCC- Heat flows from hot object to a colder one; the - Heat flows from hot object to a colder one; the

reverse never happens spontaneouslyreverse never happens spontaneously- Gas expands into an evacuated bulb - Gas expands into an evacuated bulb

spontaneously; the reverse process (gas spontaneously; the reverse process (gas molecules gathering into one bulb) is not molecules gathering into one bulb) is not spontaneousspontaneous

OYO’sOYO’s13.10 Which of the following processes are 13.10 Which of the following processes are

spontaneous and which are spontaneous and which are nonspontaneous?nonspontaneous?a. Dissolving NaCl in soupa. Dissolving NaCl in soupb. Climbing Mt. Everestb. Climbing Mt. Everestc. Spreading fragrance in a room by c. Spreading fragrance in a room by removing the bottle capremoving the bottle capd. Separating He and Ne from a mixture d. Separating He and Ne from a mixture of gasesof gases

VIII. Gibbs Free NRG (VIII. Gibbs Free NRG (GG – energy – energy available to do work): How to tell if available to do work): How to tell if a reaction is spontaneous or nota reaction is spontaneous or not

A. Two driving forces determine whether a A. Two driving forces determine whether a reaction is spontaneous or notreaction is spontaneous or not

1. Whether the product(s) have more 1. Whether the product(s) have more entropy than the reactant(s)entropy than the reactant(s)

2. Whether the reaction is endothermic or 2. Whether the reaction is endothermic or exothermicexothermica. Exothermic a. Exothermic increasesincreases the entropy the entropy

of of the surroundingsthe surroundingsb. Endothermic b. Endothermic decreases decreases the the

entropy entropy of the surroundingsof the surroundings

B. The Gibbs Free Energy equation puts B. The Gibbs Free Energy equation puts them togetherthem together

1. The equation: ∆1. The equation: ∆GG = ∆ = ∆HH - T∆ - T∆SS

2. When ∆2. When ∆GG < < 0, reaction is 0, reaction is spontaneousspontaneous

3. When ∆3. When ∆GG > 0, reaction is > 0, reaction is notnot spontaneousspontaneous

4. Can also use: 4. Can also use:

∆∆GGrxnrxn oo = ∑ = ∑ ∆∆GGffoo (prod.) - (prod.) - ∑ ∑ ∆∆GGff

oo (reac.) (reac.)

C. Factors Affecting the sign in ∆C. Factors Affecting the sign in ∆GG

D. Using ∆D. Using ∆GG = ∆ = ∆HH - T∆ - T∆SS

1. 2NO(g) + O1. 2NO(g) + O22(g) → 2NO(g) → 2NO22(g) (g)

∆∆HH = -114.2 kJ = -114.2 kJ∆∆SS = -147 J/mol K = -147 J/mol K

*What is ∆*What is ∆GG at 532 at 532ooC (805 K)?C (805 K)?*Is it spontaneous at that *Is it spontaneous at that

temperature?temperature?*Is there a temperature at which this *Is there a temperature at which this

rxn rxn is spontaneous?is spontaneous?

2. What is the ∆2. What is the ∆GG for the following rxn at for the following rxn at 298 K? Is it spontaneous?298 K? Is it spontaneous?

Ca(s) + 2HCl(aq) → CaClCa(s) + 2HCl(aq) → CaCl22(s) + H(s) + H22(g)(g)

OYO’sOYO’s

13.11 The ∆13.11 The ∆HHoo of a certain chemical reaction is of a certain chemical reaction is 10.0 kJ whereas ∆10.0 kJ whereas ∆SSoo is 123 J/K. At what is 123 J/K. At what temperatures is this reaction spontaneous?temperatures is this reaction spontaneous?

13.12 Is the following synthesis of ethyl alcohol a 13.12 Is the following synthesis of ethyl alcohol a spontaneous reaction at 298 K?spontaneous reaction at 298 K?

2CH2CH44(g) + O(g) + O22(g) (g) →→ C C22HH66O(l) + HO(l) + H22O(l)O(l)